Mechanics of Materials: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

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1Formulas and methods

Axial stress & strain

Normal stress: σ = F/A
Strain: ε = ΔL/L
Hooke’s law: σ = E·ε
Elongation: ΔL = FL/(EA)
Thermal strain: ε_T = α·ΔT

Torsion of shafts

Shear stress: τ = T·r/J
Solid shaft: J = πd⁴/32
Hollow shaft: J = π(D⁴ − d⁴)/32
Angle of twist: φ = TL/(GJ)
Power: P = T·ω = 2πnT/60

Method

Find the polar moment of area: J = πd⁴/32 for a solid shaft.
Shear stress at the surface: τ = T·r/J, with r the outer radius.
Angle of twist: φ = TL/(GJ) in radians.
If power is given, find the torque first with P = T·ω, ω = 2πn/60.

Shear force & bending moment

Shear and moment: V = ΣF on one side, M = ΣF·d on one side
Bending stress: σ = M·y/I
Rectangle: I = bh³/12
Section modulus: W = I/y_max
Simply supported, uniform load: M_max = wL²/8

Method

Find the support reactions from equilibrium of the whole beam.
Cut the beam at the section you need and keep one side.
Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
For the maximum moment, find where the shear force is zero.

Centroid & second moment of area

Centroid: ȳ = ΣA_iy_i/ΣA_i
Rectangle: I = bh³/12
Parallel-axis theorem: I = I_c + A·d²
Composite section: I = Σ(I_c,i + A_id_i²)

Mohr’s circle & stress transformation

Centre: C = (σ_x + σ_y)/2
Radius: R = √(((σ_x − σ_y)/2)² + τ_xy²)
Principal stresses: σ_1,2 = C ± R
Principal direction: tan 2θ_p = 2τ_xy/(σ_x − σ_y)
Inclined plane: σ_θ = C + (σ_x − σ_y)/2·cos 2θ + τ_xy·sin 2θ

Method

Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
The principal stresses are C ± R and the largest in-plane shear stress is R.
Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).

Beam deflection

Point load at midspan: δ = PL³/(48EI)
Uniform load, simply supported: δ = 5wL⁴/(384EI)
Cantilever, end load: δ = PL³/(3EI)
Cantilever, uniform load: δ = wL⁴/(8EI)
Superposition: δ = δ₁ + δ₂

Column buckling

Euler load: P_cr = π²EI/(KL)²
Effective length factor: K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress: σ_cr = P_cr/A
Slenderness: λ = KL/r, r = √(I/A)

2Practice problems

1.Axial stress & strainMedium

A bar is made of a steel segment (diameter 40 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 30 mm, length 500 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.

segment 1: steel, Ø40 mm, 500 mm, E = 200 GPasegment 2: copper, Ø30 mm, 500 mm, E = 120 GPaδ = 0.4 mm

P = kN

2.Axial stress & strainHard

A steel segment (diameter 25 mm, length 400 mm, E = 200 GPa) and a segment made of aluminium (diameter 60 mm, length 200 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø25 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø60 mm, 200 mm, E = 70 GPaP = 20 kN

σ₂ = MPa

3.Torsion of shaftsMedium

A hollow steel shaft (outer diameter 80 mm, inner diameter 48 mm, length 1.5 m) is fixed at one end and twisted by a torque of 1500 N·m at the other. Find the angle of twist φ in degrees.

T = 1500 N·md_o = 80 mmd_i = 48 mmL = 1.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid brass shaft transmits 30 kW at 500 rpm. The shear stress must not exceed 40 MPa and the twist must not exceed 0.5° per m of length. Find the smallest diameter d that satisfies both limits.

P = 30 kWN = 500 rpmτ_allow = 40 MPaθ_allow = 0.5°/mG = 39 GPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 3 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 8 mw = 3 kN/m from 0 to 8 mx = 3 m

M_x = kN·m

6.Shear force & bending momentHard

Find the bending moment M_x at the section x = 3 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mP = 40 kN at 7 mw = 6 kN/m from 0 to 10 mx = 3 m

M_x = kN·m

7.Centroid & second moment of areaMedium

A channel section lies with its open side up: a base plate 260 mm wide with two upright legs, 190 mm high overall. Base and legs are all 8 mm thick. Find its second moment of area I_x about the horizontal axis x through its centroid.

b = 260 mmh = 190 mmt = 8 mm

I_x = × 10⁶ mm⁴

8.Centroid & second moment of areaHard

A 150 mm × 150 mm plate has a rectangular cut-out 80 mm wide and 20 mm high, centred on the vertical centre line with its centre 45 mm above the bottom. Find the second moment of area I_x about the horizontal axis through the centroid.

b = 150 mmh = 150 mmcut-out: 80 × 20 mmhole centre: y = 45 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −20 MPa, σᵧ = 20 MPa and τₓᵧ = 50 MPa (tension positive). Find the angle θ_p from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σ_x = −20 MPaσ_y = 20 MPaτ_xy = 50 MPa

θ_p = °

10.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 500 mm and wall thickness 12 mm holds gas at a gauge pressure of 3 MPa. It is made from plate welded along a helical seam at 35° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 500 mmt = 12 mmp = 3 MPaα = 35°

σ_n = MPa

11.Beam deflectionMedium

A simply supported beam carries a point load P at a = 2 m from support A. Find the deflection δ_P under the load.

P = 12 kNa = 2 mL = 5 mE = 70 GPa (aluminium)I = 84 × 10⁶ mm⁴

δ_P = mm

12.Beam deflectionHard

A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the reaction R_B at the prop.

L = 6 mw = 5 kN/m

R_B = kN

13.Column bucklingMedium

A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 4 mE = 70 GPa (aluminium)d = 80 mm

P_cr = kN

14.Column bucklingHard

A column of length L made of aluminium is fixed at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 4 mE = 70 GPa (aluminium)d = 25 mm

σ_cr = MPa

3Solutions

1.Axial stress & strain

Answer: P = 50.74 kN

Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, P_i = P·E_iA_i/ΣEA, and σ_i = E_i·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.

Areas

A₁ = 1257 mm², A₂ = 706.9 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·1257) = 1.989 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 500/(120000·706.9) = 5.895 × 10⁻⁶ mm/N

Load

P = 0.4/7.884 × 10⁻⁶ = 50735 N = 50.74 kN

Answer

P = 50.74 kN

2.Axial stress & strain

Answer: σ₂ = 5.668 MPa

Stiffnesses

k₁ = 200000·490.9/400 = 245437 N/mm

k₂ = 70000·2827/200 = 989602 N/mm

Share of segment 2

F₂ = 20000·989602/1.235 × 10⁶ = 16025 N

Stress

σ₂ = 16025/2827 = 5.668 MPa (compression)

Answer

σ₂ = 5.668 MPa

3.Torsion of shafts

Answer: φ = 0.4604 °

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τ_max = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(d_o⁴ − d_i⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.

Polar second moment of area

J = π(80⁴ − 48⁴)/32 = 3.5 × 10⁶ mm⁴

Angle of twist

T = 1500 N·m = 1.5 × 10⁶ N·mm

L = 1.5 m = 1500 mm

G = 80 GPa = 80000 MPa

φ = TL/(GJ) = 1.5 × 10⁶·1500/(80000·3.5 × 10⁶) = 0.008036 rad

φ = 0.008036·180/π = 0.4604°

Answer

φ = 0.4604°

4.Torsion of shafts

Answer: d = 64.35 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 30000/52.36 = 573 N·m

Diameter for strength

T = 573 N·m = 572958 N·mm

d_τ = ∛(16·572958/(π·40)) = 41.78 mm

Diameter for stiffness

θ = 0.5·π/180/1000 = 8.727 × 10⁻⁶ rad/mm

G = 39 GPa = 39000 MPa

d_φ = (32·572958/(π·39000·8.727 × 10⁻⁶))^1/4 = 64.35 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(41.78, 64.35) = 64.35 mm

Answer

d = 64.35 mm

Here stiffness governs.

5.Shear force & bending moment

Answer: M_x = 22.5 kN·m

Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 3·8 = 24 kN, acting at x̄ = 4 m

ΣM_A = 0: R_B·8 = 3·8·4

R_B = 96/8 = 12 kN

ΣF_y = 0: R_A = 24 − 12 = 12 kN

Cut at x = 3 m and keep the left part

Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.

M = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 12·3 − 3·3·1.5 = 22.5 kN·m

Answer

M_x = 22.5 kN·m

6.Shear force & bending moment

Answer: M_x = 99 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 6·10 = 60 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 40·7 + 6·10·5

R_B = 580/10 = 58 kN

ΣF_y = 0: R_A = 100 − 58 = 42 kN

Cut at x = 3 m and keep the left part

Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.

M = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 42·3 − 6·3·1.5 = 99 kN·m

Answer

M_x = 99 kN·m

7.Centroid & second moment of area

Answer: I_x = 19 × 10⁶ mm⁴

Composite areas: split the section into rectangles. The centroid is the area-weighted average ȳ = ΣAᵢyᵢ/ΣAᵢ. The second moment of area about the centroidal axis adds each part’s own bh³/12 and its parallel-axis term A·d², where d is the distance from the part’s centroid to the overall centroid. Bending stress is σ = M·y/I.

Split into rectangles

Take the base at full width and the two legs above it (height h − t), so nothing is counted twice.

base: A₁ = 260·8 = 2080 mm², y₁ = 4 mm

two legs: A₂ = 2·8·182 = 2912 mm², y₂ = 8 + 182/2 = 99 mm

Centroid

ȳ = (2080·4 + 2912·99)/4992 = 59.42 mm

Parallel-axis theorem

base: 260·8³/12 + 2080·(4 − 59.42)² = 6.399 × 10⁶ mm⁴

two legs: 2·8·182³/12 + 2912·(99 − 59.42)² = 1.26 × 10⁷ mm⁴

I_x = 6.399 × 10⁶ + 1.26 × 10⁷

I_x = 1.9 × 10⁷ mm⁴ = 19 × 10⁶ mm⁴

Answer

I_x = 19 × 10⁶ mm⁴

8.Centroid & second moment of area

Answer: I_x = 40.58 × 10⁶ mm⁴

Sections with holes: treat a hole as a negative area. The centroid is ȳ = (A₁y₁ − A₂y₂)/(A₁ − A₂), with A₁, y₁ for the full plate and A₂, y₂ for the hole. For the second moment of area about the centroidal axis, use the parallel-axis theorem for each part and subtract the hole: I = (I_c1 + A₁d₁²) − (I_c2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has I_c = bh³/12 and a circle I_c = πd⁴/64.

Centroid

ȳ = (22500·75 − 1600·45)/(22500 − 1600) = 77.3 mm

Full plate about ȳ

150·150³/12 + 22500·−2.297² = 42.31 × 10⁶ mm⁴

Hole about ȳ

80·20³/12 + 1600·−32.3² = 1.722 × 10⁶ mm⁴

Subtract

I_x = 40.58 × 10⁶ mm⁴

Answer

I_x = 40.58 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: θ_p = 55.9 °

Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θ_p = atan2(2·50, −20 − 20) = 111.8°

Angle on the element

θ_p = 55.9°

Answer

θ_p = 55.9°

Positive angles are counter-clockwise.

10.Mohr’s circle & stress transformation

Answer: σ_n = 52.22 MPa

Wall stresses

r = 500/2 = 250 mm

σ_h = pr/t = 3·250/12 = 62.5 MPa

σ_a = pr/(2t) = 31.25 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

σ_n = 31.25·sin²35° + 62.5·cos²35° = 52.22 MPa

Answer

σ_n = 52.22 MPa

11.Beam deflection

Answer: δ_P = 4.898 mm

Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition).

Consistent units

Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.

P = 12000 N, a = 2000 mm, b = 3000 mm, L = 5000 mm

EI = 5.88 × 10¹² N·mm²

Deflection under the load

δ_P = Pa²b²/(3EIL) = 12000·2000²·3000²/(3·5.88 × 10¹²·5000) = 4.898 mm

Answer

δ_P = 4.898 mm

12.Beam deflection

Answer: R_B = 11.25 kN

Remove the prop

Without the prop the beam is a cantilever. Its free end would deflect under the load.

δ_load = wL⁴/(8EI)

Compatibility at the prop

The prop force R_B alone lifts the free end by R_BL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.

R_B·L³/(3EI) = wL⁴/(8EI) → R_B = 3wL/8 = 3·5·6/8 = 11.25 kN

Answer

R_B = 11.25 kN

13.Column buckling

Answer: P_cr = 177.2 kN

Euler buckling: a slender column becomes unstable at P_cr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 0.7·4000 = 2800 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/2800² = 177179 N = 177.2 kN

Answer

P_cr = 177.2 kN

14.Column buckling

Answer: σ_cr = 6.747 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·25⁴/64 = 19175 mm⁴

Radius of gyration and slenderness

A = 490.9 mm²

r = √(I/A) = √(19175/490.9) = 6.25 mm

KL = 0.5·4000 = 2000 mm

KL/r = 2000/6.25 = 320

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/320² = 6.747 MPa

Answer

σ_cr = 6.747 MPa

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